2010 AMC 12B Problem 9

Below is the professionally curated solution for Problem 9 of the 2010 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12B solutions, or check the answer key.

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Concepts:prime factorization perfect power least common multiple

Difficulty rating: 1520

Let $n$ be the smallest positive integer such that $n$ is divisible by $20,$ $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n?$

$3$

$4$

$5$

$6$

$7$

Solution:

To be smallest, $n$ uses only the primes of $20,$ so $n=2^a\cdot5^b$ with $a\ge2$ and $b\ge1.$

Since $n^2=2^{2a}5^{2b}$ is a perfect cube, $3\mid a$ and $3\mid b.$ Since $n^3=2^{3a}5^{3b}$ is a perfect square, $2\mid a$ and $2\mid b.$ Hence $6\mid a$ and $6\mid b.$

The smallest choice is $a=b=6,$ so $n=2^6\cdot5^6=10^6=1{,}000{,}000,$ which has $7$ digits.

Thus, the correct answer is E.

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